In a right-angled triangle ∆ABC, the altitude AB is 5 cm, and the base BC is 12 cm. P and Q are two points on BC such that the areas of ∆ABP, ∆ABQ and ∆ABC are in arithmetic progression. If the area of ∆ABC is 1.5 times the area of ∆ABP, the length of PQ, in cm, is
Explanation:
Given, AB = 5 and BC = 12
Area of ∆ABP = ½ × AB × BP = 2.5 × BP …(1) Area of ∆ABQ = ½ × AB × BQ = 2.5 × BQ …(2) Area of ∆ABC = ½ × AB × BC = 2.5 × BC …(3)
Given, (3) = 1.5 × (1) ⇒ 2.5 BC = 1.5 × 2.5 × BP ⇒ BC = 1.5 × BP ⇒ 12 = 1.5 × BP ⇒ BP = 8
Also, Areas of ∆ABP, ∆ABQ and ABC are in arithmetic progression. ⇒ 2 × (2) = (1) + (3) ⇒ 2 × 2.5 × BQ = 2.5 × BP + 2.5 × BC ⇒ 2BQ = BP + BC ⇒ 2BQ = 8 + 12 ⇒ BQ = 10
∴ PQ = BQ – BP = 10 – 8 = 2
Hence, 2.
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